Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $k \neq 0$. $x = \dfrac{-2k + 10}{-8k - 24} \div \dfrac{k^2 - 7k + 10}{-2k + 4} $
Dividing by an expression is the same as multiplying by its inverse. $x = \dfrac{-2k + 10}{-8k - 24} \times \dfrac{-2k + 4}{k^2 - 7k + 10} $ First factor the quadratic. $x = \dfrac{-2k + 10}{-8k - 24} \times \dfrac{-2k + 4}{(k - 5)(k - 2)} $ Then factor out any other terms. $x = \dfrac{-2(k - 5)}{-8(k + 3)} \times \dfrac{-2(k - 2)}{(k - 5)(k - 2)} $ Then multiply the two numerators and multiply the two denominators. $x = \dfrac{ -2(k - 5) \times -2(k - 2) } { -8(k + 3) \times (k - 5)(k - 2) } $ $x = \dfrac{ 4(k - 5)(k - 2)}{ -8(k + 3)(k - 5)(k - 2)} $ Notice that $(k - 2)$ and $(k - 5)$ appear in both the numerator and denominator so we can cancel them. $x = \dfrac{ 4\cancel{(k - 5)}(k - 2)}{ -8(k + 3)\cancel{(k - 5)}(k - 2)} $ We are dividing by $k - 5$ , so $k - 5 \neq 0$ Therefore, $k \neq 5$ $x = \dfrac{ 4\cancel{(k - 5)}\cancel{(k - 2)}}{ -8(k + 3)\cancel{(k - 5)}\cancel{(k - 2)}} $ We are dividing by $k - 2$ , so $k - 2 \neq 0$ Therefore, $k \neq 2$ $x = \dfrac{4}{-8(k + 3)} $ $x = \dfrac{-1}{2(k + 3)} ; \space k \neq 5 ; \space k \neq 2 $